I think it's true that $f(z)={\rm Det}(I+zT)$ can be written as $\prod\_n(1+z\lambda\_n)$ where $\lambda\_n$ are the eigenvalues of $T$ (I just say think, because I haven't really looked at Fredholm determinants before). Then, if $f$ is nowhere zero, we have $\lambda\_n=0$, so $f(z)=1$. Therefore, the answer to the question is no.
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George LowtherOct 13 '11 at 0:36

2 Answers
2

Here's a proof that $\exp(z)$ is not a characteristic function using the product expansion for the determinant, which is essentially equivalent to Lidskii's theorem stating that the trace of a trace class operator is the sum of its eigenvalues.

If $T$ is a trace class operator on a Hilbert space $\mathcal{H}$ with eigenvalues $\lambda_n$ (counted up to algebraic multiplicity), then $\sum_n\vert\lambda_n\vert < \infty$ and
$$
{\rm det}(1+zT)\equiv\sum_{k=0}^\infty{\rm Tr}(\Lambda^kT)z^k=\prod_n(1+z\lambda_n)
$$
for all $z\in\mathbb{C}$.

This is not an answer, but just a sketch of a possible answer; I think the answer is no.
For $T$ as desired, consider $Det(I-z^2T^2)=Det(I-zT)\cdot Det(I+zT)=e^{-z}\cdot e^z\equiv 1$. Considering the Taylor expansion of this function, one can conclude that $Trace(T^{2n})=0$ for any natural $n$. I think this yields $T^2=0$, and then the claim follows.

Actually, I am not sure if such determinant may have positive exponential type - this could be another approach.

If $T^2$ is a quasinilpotent trace class operator, then it has trace zero (Lidskii). So all powers of $T^2$ have trace zero. Does this imply that $\Tr(\wedge^{2k} T) \neq 1/(2k!)$?
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Yemon ChoiOct 12 '11 at 22:07

If so, maybe my first argument that $T^2=0$ is wrong; I got it from the finite-dimensional situation.
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kap44Oct 12 '11 at 22:19

Ok. That proves that $T$ is quasinilpotent (all eigenvalues are zero). Then, I think, $\Lambda^nT$ is also quasinilpotent and ${\rm Det}(I+zT)=1$, giving a negative result to the question.
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George LowtherOct 12 '11 at 22:53